lc498. 对角线遍历（模拟）

给你一个大小为 m x n 的矩阵 mat ，请以对角线遍历的顺序，用一个数组返回这个矩阵中的所有元素。

示例 1：

输入：mat = [[1,2,3],[4,5,6],[7,8,9]]
输出：[1,2,4,7,5,3,6,8,9]
示例 2：

输入：mat = [[1,2],[3,4]]
输出：[1,2,3,4]

提示：

m == mat.length
n == mat[i].length
1 <= m, n <= 104
1 <= m * n <= 104
-105 <= mat[i][j] <= 105

class Solution {
    public int[] findDiagonalOrder(int[][] mat) {
        int n = mat.length, m = mat[0].length;
        int[] res = new int[n * m];
        int idx = 0;
        //idr = 1 表示向右上方移动 -1表示左下方
        int idr = 1, x = 0, y = 0;
        while (idx < n * m) {
            res[idx ++] = mat[x][y];
            int nx = x, ny = y;
            if (idr == 1) {
                nx = nx - 1;
                ny = ny + 1;
            } else {
                nx = nx + 1;
                ny = ny - 1;
            }
            //超出界限
            if (idx < n * m && (nx < 0 || nx >= n || ny < 0 || ny >= m)) {
                if (idr == 1) {
                    nx = y + 1 < m ? x : x + 1;
                    ny = y + 1 < m ? y + 1 : y;
                } else {
                    nx = x + 1 < n ? x + 1 : x;
                    ny = x + 1 < n ? y : y + 1;
                }
                idr *= -1;
            }
            x = nx; y = ny;
        }
        return res;
    }
}